
The amount of work done in blowing a soup bubble such that its diameter increases from $d\,to\,D$ is: (surface tension of solution is T)
(A) $4\pi ({D^2} - {d^2})T$
(B) $8\pi ({D^2} - {d^2})T$\[\]
(C) $\pi ({D^2} - {d^2})$
(D) $2\pi ({D^2} - {d^2})T$
Answer
163.8k+ views
Hint: First start to find out what will be the relation between the amount of work done in blowing a soup bubble, its diameter and the surface tension then try to put all the information provided in the question in that relation. Also the D is greater than d as given in the question so put its value accordingly.
Formula used:
$W=T\times \Delta A$
W is the work done, T is the surface tension and $\Delta A$is the change in area
Complete answer:
Start with the formula of the amount of work done in blowing a soap bubble. We know that, Work done is equal to tension in surface energy and surface energy is tension multiply to change in area as follows:
$W = T \times \Delta A$
Where,
W is work done
T is surface tension and
$\Delta A$is the change in area.
Now, in given case of soap bubble, work done will be;
$W = 8\pi T(r_2^2 - r_2^2)$ (equation 1)
From the question, we know that;
${r_1} = \dfrac{d}{2}\,and\,{r_2} = \dfrac{D}{2}$
Putting the value of \[{r_{1\,}}\,and\,{r_2}\] in equation 1, we get;
\[W = 8\pi T\{ (\dfrac{{{D^2}}}{4}) - (\dfrac{{{d^2}}}{4})\} \]
After solving, we get;
\[W = 2\pi T({D^2} - {d^2})\]
Hence the correct answer is Option D
Note: The thing to be kept in mind while talking about the surface tension. Fluids have a feature called surface tension that causes the fluids to tend to minimize or shrink their surface area. The force acting along the length of a fluid's surface is known as surface tension.
Formula used:
$W=T\times \Delta A$
W is the work done, T is the surface tension and $\Delta A$is the change in area
Complete answer:
Start with the formula of the amount of work done in blowing a soap bubble. We know that, Work done is equal to tension in surface energy and surface energy is tension multiply to change in area as follows:
$W = T \times \Delta A$
Where,
W is work done
T is surface tension and
$\Delta A$is the change in area.
Now, in given case of soap bubble, work done will be;
$W = 8\pi T(r_2^2 - r_2^2)$ (equation 1)
From the question, we know that;
${r_1} = \dfrac{d}{2}\,and\,{r_2} = \dfrac{D}{2}$
Putting the value of \[{r_{1\,}}\,and\,{r_2}\] in equation 1, we get;
\[W = 8\pi T\{ (\dfrac{{{D^2}}}{4}) - (\dfrac{{{d^2}}}{4})\} \]
After solving, we get;
\[W = 2\pi T({D^2} - {d^2})\]
Hence the correct answer is Option D
Note: The thing to be kept in mind while talking about the surface tension. Fluids have a feature called surface tension that causes the fluids to tend to minimize or shrink their surface area. The force acting along the length of a fluid's surface is known as surface tension.
Recently Updated Pages
Fluid Pressure - Important Concepts and Tips for JEE

JEE Main 2023 (February 1st Shift 2) Physics Question Paper with Answer Key

Impulse Momentum Theorem Important Concepts and Tips for JEE

Graphical Methods of Vector Addition - Important Concepts for JEE

JEE Main 2022 (July 29th Shift 1) Chemistry Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 1) Physics Question Paper with Answer Key

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Charging and Discharging of Capacitor

Other Pages
NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement
